What is the derivative of (2x^2+4x+2/(sqrt(x))) ?

The derivative of (2x^2+4x+2/(sqrt(x))) is (3x^2+2x-1)/(xsqrt(x))

What is the first derivative of (2x^2+4x+2/(sqrt(x))) ?

The first derivative of (2x^2+4x+2/(sqrt(x))) is (3x^2+2x-1)/(xsqrt(x))

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\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 2}{x})

\frac{3 x ^{2} + 2 x - 1}{x x}

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 2}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 2 x ^{2} + 4 x + 2 ) x - \frac{d}{d x} ( x ) ( 2 x ^{2} + 4 x + 2 )}{( x ) ^{2}}

\frac{d}{d x} (2 x^{2} + 4 x + 2) = 4 x + 4

\frac{d}{d x} (x) = \frac{1}{2 x}

= \frac{( 4 x + 4 ) x - \frac{1}{2 x} ( 2 x ^{2} + 4 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 4 x + 4 ) x - \frac{1}{2 x} ( 2 x ^{2} + 4 x + 2 )}{( x ) ^{2}} : \frac{3 x ^{2} + 2 x - 1}{x x}

= \frac{3 x ^{2} + 2 x - 1}{x x}

\frac{d y}{d x} = \frac{- y ^{2}}{1 + y ^{2}} e^{- x}

x \to 2 - lim (\frac{x ^{3} + 1}{x - 2})

\int \frac{1}{( x + 5 ) ^{2} ( x - 3 )} d x

\frac{\partial}{\partial y} (4 x y - x^{4} - y^{4} + 1)

n = 2 \sum \infty \frac{( 3 x - 2 ) ^{n}}{n}

What is the derivative of (2x^2+4x+2/(sqrt(x))) ?
The derivative of (2x^2+4x+2/(sqrt(x))) is (3x^2+2x-1)/(xsqrt(x))
What is the first derivative of (2x^2+4x+2/(sqrt(x))) ?
The first derivative of (2x^2+4x+2/(sqrt(x))) is (3x^2+2x-1)/(xsqrt(x))